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Question

A small block of mass m3=2kg of negligible size is placed on a plank of mass m2=4kg and length 4m which is further placed on a frictionless surface as shown in figure. If m1=8kg and system starts from rest,

A
Time after which m3 falls from m2 is
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B
Time after which m3 falls from m2 is 2 sec
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C
Velocity of m1 at 2 sec is 12 m/s
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D
Velocity of m2 at t = 2 sec is 11.4 m/s
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Solution

The correct options are
B Time after which m3 falls from m2 is 2 sec
C Velocity of m1 at 2 sec is 12 m/s
Assume all the blocks move together, then a=m1gm1+m2+m3
=808+4+2=8014=5.7 m
Friction acting on m3=m3a=2×5.7
=11.4 N
But maximum friction on m3 can only be, f=m3gμ=20×0.4=8N
Therefore, m3 sliding on m2, and f=8N


a=m1gfm1+m2=80812=6 m/s2

acceleration (a0) of m3 is a0=fm3=82=4
Relative acceleration of m3 with respect to m2 is 64=2m/s2
The distance it has to cover is 4m.
s=ut+12at2
4=12×2×t2
t=2 sec
Velocity of m1 and m2 at t = 2 sec is u+at
=0+6×2=12 m/s

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